A) \[3\]
B) \[6\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
Since, the planes \[2x-2y+z+3=0\] and \[2x-2y+z+\frac{5}{2}=0\] are parallel to each other. \[\therefore \] Distance between them \[=\frac{|{{c}_{2}}-{{c}_{1}}|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}\] \[=\frac{\left| \frac{5}{2}-3 \right|}{\sqrt{4+4+1}}\] \[=\frac{\frac{1}{2}}{3}=\frac{1}{6}\]
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